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# NEW!!! Tough and tricky exponents and roots questions

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 15:46
7
4
7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

Must know for the GMAT: if $$x$$ is a positive integer then $$\sqrt{x}$$ is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether $$x$$ is a perfect square.

(1) $$\sqrt{7*x}$$ is an integer --> $$x$$ can not be a perfect square because if it is, for example if $$x=n^2$$ for some positive integer $$n$$ then $$\sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}$$. Sufficient.

(2) $$\sqrt{9*x}$$ is not an integer --> $$\sqrt{9*x}=3*\sqrt{x}\neq{integer}$$ --> $$\sqrt{x}\neq{integer}$$. Sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 15:49
7
9
8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

(1) $$x^6+y^9=0$$ --> $$(x^2)^3=(-y3)^3$$ ---> $$x^2=-y^3$$ --> $$x^2+y^3=0$$. Sufficient.

(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$ --> $$3^{3x^2}=\frac{3}{3^{3y^2}*3}$$ --> $$3^{3x^2}*3^{3y^2}=1$$ --> $$3^{3x^2+3y^2}=1$$ --> $$3x^2+3y^2=0$$ (the power of 3 must be zero in order this equation to hold true) --> $$x^2+y^2=0$$ the sum of two non-negative values is zero --> both $$x$$ and $$y$$ must be zero --> $$x=y=0$$ --> $$x^2+y^3=0$$. Sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 15:52
13
10
9. If $$x$$, $$y$$ and $$z$$ are non-zero numbers, what is the value of $$\frac{x^3+y^3+z^3}{xyz}$$?
(1) $$xyz=-6$$
(2) $$x+y+z=0$$

(1) $$xyz=-6$$ --> infinitely many combinations of $$x$$, $$y$$ and $$z$$ are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.

(2) $$x+y+z=0$$ --> $$x=-(y+z)$$ --> substitute this value of x into the expression in the stem --> $$\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}$$, as $$x=-(y+z)$$ then: $$\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3$$. Sufficient.

Must know for the GMAT: $$(x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3$$ and $$(x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3$$.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 15:54
11
9
10. If $$x$$ and $$y$$ are non-negative integers and $$x+y>0$$ is $$(x+y)^{xy}$$ an even integer?
(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$
(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$

(1) $$2^{x-y}=\sqrt[(x+y)]{16}$$ --> $$2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}$$ --> equate the powers: $$x-y=\frac{4}{x+y}$$ --> $$(x-y)(x+y)=4$$.

Since both $$x$$ and $$y$$ are integers (and $$x+y>0$$) then $$x-y=2$$ and $$x+y=2$$ --> $$x=2$$ and $$y=0$$ --> $$(x+y)^{xy}=2^0=1=odd$$, so the answer to the question is No. Sufficient. (Note that $$x-y=1$$ and $$x+y=4$$ --> $$x=2.5$$ and $$y=1.5$$ is not a valid scenario (solution) as both unknowns must be integers)

(2) $$2^x+3^y=\sqrt[(x+y)]{25}$$ --> obviously $$\sqrt[(x+y)]{25}$$ must be an integer (since $$2^x+3^y=integer$$) and as $$x+y=integer$$ then the only solution is $$\sqrt[(x+y)]{25}=\sqrt[2]{25}=5$$ --> $$x+y=2$$. So, $$2^x+3^y=5$$ --> two scenarios are possible:
A. $$x=2$$ and $$y=0$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^2+3^0=5$$, and in this case: $$(x+y)^{xy}=2^0=1=odd$$;
B. $$x=1$$ and $$y=1$$ (notice that $$x+y=2$$ holds true) --> $$2^x+3^y=2^1+3^1=5$$, and in this case: $$(x+y)^{xy}=2^1=2=even$$.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 15:56
13
16
11. What is the value of $$xy$$?
(1) $$3^x*5^y=75$$
(2) $$3^{(x-1)(y-2)}=1$$

Notice that we are not told that the $$x$$ and $$y$$ are integers.

(1) $$3^x*5^y=75$$ --> if $$x$$ and $$y$$ are integers then as $$75=3^1*5^2$$ then $$x=1$$ and $$y=2$$ BUT if they are not, then for any value of $$x$$ there will exist some non-integer $$y$$ to satisfy given expression and vise-versa (for example if $$y=1$$ then $$3^x*5^y=3^x*5=75$$ --> $$3^x=25$$ --> $$x=some \ irrational \ #\approx{2.9}$$). Not sufficient.

(2) $$5^{(x-1)(y-2)}=1$$ --> $$(x-1)(y-2)=0$$ --> either $$x=1$$ and $$y$$ is ANY number (including 2) or $$y=2$$ and $$x$$ is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) $$x=1$$ then from (1) $$3^x*5^y=3*5^y=75$$ --> $$y=2$$ and if from (2) $$y=2$$ then from (1) $$3^x*5^y=3^x*25=75$$ --> $$x=1$$. Thus $$x=1$$ and $$y=2$$. Sufficient.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 19:16
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ and integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> $$xyz=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and $$z=3^3$$ =odd perfect square then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3$$ =odd perfect square then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z}\neq{integer}$$ then $$z\neq{1}$$, therefore it must be $$3^3$$ (from 1) --> $$x$$ is a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2
(2^2)(3^2)
(2^2)(5^2)
(2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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15 Jan 2012, 02:39
IanStewart wrote:
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ and integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> $$xyz=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and $$z=3^3$$ =odd perfect square then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3$$ =odd perfect square then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z}\neq{integer}$$ then $$z\neq{1}$$, therefore it must be $$3^3$$ (from 1) --> $$x$$ is a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2
(2^2)(3^2)
(2^2)(5^2)
(2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights).

Yes, I know Ian. For (1) I just discussed two possible scenarios to get an YES and NO answers to discard this statement, rather than listing all possibilities as you did. Thank you though for elaborating more on other cases.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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17 Mar 2012, 06:28
Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

SonyGmat wrote:
Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt[3]{y})*(z^{-2})<0$$?
(1) $$\sqrt[5]{y}>\sqrt[4]{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

if we can find the sign of y we can answer the question.

(1) $$\sqrt[5]{y}>\sqrt[4]{x^2}>0$$ since it is$$[\sqrt[4]{x^2}]=[\sqrt[]{x}]$$

Sufficient

(2) $$y^3>\frac{1}{z{^4}}>0$$ since $$z^4>0$$

Sufficient

Therefore, D
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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17 Mar 2012, 06:33
1
devinawilliam83 wrote:
Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, $$\sqrt{25}=+5$$ and $$-\sqrt{25}=-5$$.

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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13 Apr 2012, 19:14
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ and integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> $$xyz=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube --> if $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3=odd \ perfect \ square$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3=odd \ perfect \ cube$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z}\neq{integer}$$ then $$z\neq{1}$$, therefore it must be $$3^3$$ (from 1) --> $$x$$ is a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Thanks
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Apr 2012, 12:30
carcass wrote:
Bunuel wrote:
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ and integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> $$xyz=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube --> if $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3=odd \ perfect \ square$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3=odd \ perfect \ cube$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z}\neq{integer}$$ then $$z\neq{1}$$, therefore it must be $$3^3$$ (from 1) --> $$x$$ is a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Thanks

Are you talking about statement 1? The values of x, y, and z analyzed there are just possible values that satisfy statement 1.

Given that: $$xyz=2^2*3^3*5^2$$ and (1) says that $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.

Now, if for example $$y=2^2*5^2=even \ perfect \ square$$ and $$z=1^3=odd \ perfect \ cube$$ then $$x=3^3$$. You can apply the similar logic to obtain other possible values of x, y, and z.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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27 Apr 2012, 11:46
Bunuel wrote:
devinawilliam83 wrote:
Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, $$\sqrt{25}=+5$$ and $$-\sqrt{25}=-5$$.

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.

but won't this leave one in a dilemma?
do i take this info with a pinch of salt, or is gmat only interested in the principal square root?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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27 Apr 2012, 11:48
nrmlvrm wrote:
Bunuel wrote:
devinawilliam83 wrote:
Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, $$\sqrt{25}=+5$$ and $$-\sqrt{25}=-5$$.

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.

but won't this leave one in a dilemma?
do i take this info with a pinch of salt, or is gmat only interested in the principal square root?

No ambiguity there whatsoever: even roots have only non-negative value on the GMAT Again: $$\sqrt{25}=5$$, NOT +5 or -5.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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05 Sep 2013, 12:30
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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05 Sep 2013, 12:35
1
Acropora wrote:
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?

23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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28 Apr 2014, 13:10
Hi,

7. If x is a positive integer is \sqrt{x} an integer?
(1) \sqrt{7*x} is an integer
(2) \sqrt{9*x} is not an integer

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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29 Apr 2014, 07:46
sheolokesh wrote:
Hi,

7. If x is a positive integer is \sqrt{x} an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$is not an integer

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..

For (1) x=7 is not the only possible value to satisfy $$\sqrt{7*x}=integer$$. For example, x can also be 7*4, 7*9, 7*16, ... basically any number of a form 7*integer^2.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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01 Jul 2014, 10:20
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards,
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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01 Jul 2014, 10:33
gkashyap wrote:
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards,
Gajendra

$$(x-y)(x+y)=23=prime$$. 23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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23 Jul 2014, 06:39
Bunuel wrote:
1. If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?
(1) $$100<y^2<x^2<169$$
(2) $$x^2-y^2=23$$

(1) $$100<y^2<x^2<169$$ --> since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> $$y=11$$ and $$x=12$$. Sufficient.(As cyclicity of units digit of $$7$$ in integer power is $$4$$, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$ --> $$(x-y)(x+y)=23=prime$$ --> since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$ --> $$y=11$$ and $$x=12$$. Sufficient.

Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!
Re: NEW!!! Tough and tricky exponents and roots questions &nbs [#permalink] 23 Jul 2014, 06:39

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